Let the first term of a geometric sequence be $\frac{3}{4}$, and let the second term be $15$. What is the smallest $n$ for which the $n$th term of the sequence is divisible by one million?
Answer: The common ratio is $$\frac{15}{\frac{3}{4}} = 20$$Therefore, the $n$th term is $(20^{n-1}) \left(\frac{3}{4}\right)$.


If one million (aka $10^6$) divides the $n$th term, then it must be divisible by $5^6$. This can only happen if $n-1$ is at least $6$, or $n \ge 7$.


The $7$th term is $$\left(20^6\right) \left(\frac{3}{4}\right) = \left(4\right)^6\left(5\right)^6\left(\frac{3}{4}\right) = (2)^{10}(5)^6(3),$$which is divisible by $(2)^6(5)^6=10^6$, so the answer is indeed $\boxed{7}$.